Problem: Let $a(x)=2x^4+7x^3+2x^2+2x-3$, and $b(x)=2x^2+x+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{2x^4+7x^3+2x^2+2x-3}{2x^2+x+1}$ : We divide ${2x^2}$ into ${2x^4}$ to get ${x^2}$ : $ \hphantom{1567|1444474} {x^2}\\ {{{2x^2}+x+1}}|\overline{{2x^4}+7x^3+2x^2+2x-\ 3}\\ \hphantom{37...8........|}\llap{-}\underline{(2x^4+\ x^3\ +\ x^2)}\\ \hphantom{37|3....998.........}+6x^3+x^2 \\ $ [What did we do here?] Next, we divide ${2x^2}$ into ${6x^3}$ to get ${+3x}$, and continue doing this until we find the quotient: $ \hphantom{1567|166664} {x^2 \ \ {+ \ \ 3x}\ -\ 1}\\ {{{2x^2}+x+1}}|\overline{2x^4+7x^3+2x^2+2x-3}\\ \hphantom{37...8........|}\llap{-}\underline{(2x^4+\ x^3\ +\ x^2)}\\ \hphantom{37|3....998.........}{+6x^3}+x^2+2x \\ \hphantom{37.......888...8.....|}\llap{-}\underline{(6x^3+3x^2+3x)}\\ \hphantom{37|3...............99999.....}{-2x^2\ -\ x\ - 3}\\ \hphantom{378888888888888888.|}\llap{-}\underline{(-2x^2-\ x\ -\ 1)}\\ \hphantom{37|3............77777777777...9999.....}{-2}\\ $ [What did we do here?] The process stops here because $2x^2+x+1$ is a polynomial of the second degree and $-2$ is a polynomial of the zeroth degree. So it follows that ${r(x)}={-2}$, ${q(x)}={x^2+3x-1}$, and $ \dfrac{2x^4+7x^3+2x^2+2x-3}{2x^2+x+1}={x^2+3x-1}+\dfrac{{-2}}{2x^2+x+1}$ To conclude, $q(x)=x^2+3x-1$ $r(x)=-2$